Problem of probability of two independent events If A and B are two independent events such that $P(AB')=\cfrac{3}{25}$ and $P(BA')=\cfrac{8}{25}$ and $P(A)<P(B)$, then what is the value of $P(A)$?
I'm getting two answers of this question. So I can't be sure whether I am correct.Please help me solving this problem. Thanks in advance.
 A: If  $A, B$ are independent events then so are $A^{\complement},B$ and $A,B^{\complement}$ so we have $2$ equalities:


*

*$P(B)-P(A)P(B)=(1-P(A))P(B)=P(A^{\complement})P(B)=P(A^{\complement}\cap B)=\frac8{25}$

*$P(A)-P(A)P(B)=P(A)(1-P(B))=P(A)P(B^{\complement})=P(A\cap B^{\complement})=\frac3{25}$
Subtraction gives $P(B)-P(A)=\frac5{25}=\frac15$
Then  substituting this in second equality gives: $P(A)-P(A)(\frac15+P(A))=P=\frac3{25}$
This leads two possibilities for $(P(A),P(B))$ but on base of $P(B)=\frac15+P(A)>P(A)$ we can pick out the right one.
A: Use the fact that $P(B) = 1-P(B')$ and by independence $P(A \cap B') = P(A) \times P(B')$: You get two unknowns in two equations. 
A: You can make a table to keep the overview:
$$\begin{array}{c|c|c|c} & A&\overline A & \\ \hline B & x & \frac8{25} & y \\ \hline \overline B& \frac{3}{25}& & \\ \hline &z&& 1\end{array}$$
If A and B are independent then the follwing to equations must hold:
$$P(B|A)=P(B)\Rightarrow \frac{z-\frac3{25}}{z}=y$$
$$P(\overline A|B)=P(\overline A)\Rightarrow \frac{\frac8{25}}{y}=1-z$$
Indeed this equation system has two solutions:
1) $y^*=\frac25, z^*=\frac15$
2) $y^*=\frac45, z^*=\frac35$
Choose the solution where $P(B)>P(A)\Rightarrow y>z$
Edit:
If I´m right, both solutions fulfill the condition. Maybe this was your problem.
